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java - 如何优先考虑在信号量中等待的线程?

转载 作者:搜寻专家 更新时间:2023-11-01 02:57:32 25 4
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我使用信号量来限制访问函数的线程数。我希望下一个被唤醒的线程应该由我将给予的一些优先级来选择,而不是默认情况下信号量唤醒它们的方式?我们怎样才能做到这一点?

这里是实现:

class MyMathUtil2 implements Runnable {
double a;
double b;
String name = "demo";
Thread t;
//static int currentCount = 0;
static int MAX_COUNT = 2;

private final Semaphore available = new Semaphore(MAX_COUNT, true);

MyMathUtil2(double v1, double v2) {
a = v1;
b = v2;
t = new Thread(this, name);
System.out.println("New thread: " + t);
t.start();
}

public void InternalPow(double a, double b) throws InterruptedException {
available.acquire();
try {
System.out.println("Power of " + a + " and " + b + " : " + Math.pow(a, b));
} finally {
available.release();
}

}

public void run() {
try {
InternalPow(a, b);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}

class TestMyMathUtil2 {
public static void main(String args[]) {
new MyMathUtil2(10.2, 8);
new MyMathUtil2(11, 56);
new MyMathUtil2(10.2, 9);
new MyMathUtil2(2, 3);
new MyMathUtil2(4, 5);
}
}

最佳答案

Semaphore 不支持优先级。

我建议使用 ThreadPoolExecutor使用 2 固定线程和 PriorityBlockingQueue来解决这个问题。

具有2 固定线程的ThreadPoolExecutor 可以确保在任何时刻,最多有2 任务在运行。其他任务将放入此 PriorityBlockingQueue,线程池将根据自定义 Comparator 从队列中检索任务。 .

这是一个例子。本例中的每个 Runnable 都应该打印一个数字。它以相反的顺序提交 Runnable:1000、999、...、1。但是 Runnable 将按自然顺序执行:1、2、....、1000 使用优先级队列。

import java.util.Comparator;
import java.util.concurrent.PriorityBlockingQueue;
import java.util.concurrent.ThreadPoolExecutor;
import java.util.concurrent.TimeUnit;

class ComparableRunnable implements Runnable {

public int index;

ComparableRunnable(int index) {
this.index = index;
}

public void run() {
System.out.println(Thread.currentThread().getName() + "-index : " + index);
try {
// sleep current thread, so the other thread can print
// this is not mandatory, without this, the result might not follow strict natural order
// for example, thread1 print 1,
// thread2 take 2 but did not print it immediatly,
// thread1 print 3,
// thread2 print 2
// the result will be 1, 3, 2,
Thread.sleep(10);
} catch (Exception e) {

}
}

public static void main(String[] args) {
int corePoolSize = 2; // fixed thread number
long ignore = 0L;

// comparator
Comparator<Runnable> comparator = new Comparator<Runnable>() {
@Override
public int compare(Runnable o1, Runnable o2) {
int index1 = ((ComparableRunnable)o1).index;
int index2 = ((ComparableRunnable)o2).index;
// you should implement this method based on your own order
return Integer.compare(index1, index2);
}
};

// use the comparator create a priority queue
PriorityBlockingQueue<Runnable> queue = new PriorityBlockingQueue<>(10, comparator);

ThreadPoolExecutor executor =
new ThreadPoolExecutor(corePoolSize, corePoolSize, ignore, TimeUnit.SECONDS, queue);

// Warm the thread pool up
// this is not mandatory, without this, it will print 1000, 999, 1, 2, ...., 998
// because the first two Runnbale will be executed once they are submitted
for (int i = 0; i < corePoolSize; i++) {
executor.execute(() -> {
try {
Thread.sleep(1000);
} catch (Exception e) {

}
});
}

// submit in 1000, 999, ..., 1 order
// print in 1, 2, 3, ..., 1000 order
for (int i = 1000; i > 0; i--) {
executor.execute(new ComparableRunnable(i));
}
}
}

结果:

pool-1-thread-1-index : 1
pool-1-thread-2-index : 2
pool-1-thread-1-index : 3
pool-1-thread-2-index : 4
pool-1-thread-2-index : 5
...
pool-1-thread-2-index : 996
pool-1-thread-2-index : 997
pool-1-thread-1-index : 998
pool-1-thread-2-index : 999
pool-1-thread-1-index : 1000

关于java - 如何优先考虑在信号量中等待的线程?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50128812/

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